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3 years ago

13

If you and 3 friends are playing tug of war and are pulling with a force of 35 N and your teammate with 37 N, while your opponen

ts are pulling with 20 N and 55 N, what is the net force and do you win?

1 answer:

mash [69]3 years ago

3 0

72N and 75N. no. hope it helps you

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Perhaps to confuse a predator, some tropical gyrinid beetles(whirligig beetles) are colored by optical interference that is duet

gulaghasi [49]

Answer:

The grating spacing of the beetle is $1.078*10^{-6}m$

Explanation:

The concept to solve this problem is relate to interference effect given in the Young's Slits. Here was demonstrated that the length of the side labelled \lambda is known as the path difference. The equation is given by,

$n\lambda = dsin\theta$

Where,

$\lambda$ = wavelenght of light

N = a positive integer: 1,2,3...

$\theta$ = Angle from the center of the wall to the dark spot

d= width of the slit

Replacing our values we have that for n=1,

$dsin\theta = n\lambda$

$dsin(30)=(1)(539*10^{-9})$

$d=1.078*10^{-6}m$

Therefore the grating spacing of the beetle is $1.078*10^{-6}m$

6 0

3 years ago

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

Which of the following is not a constant or control for this experiment?

SCORPION-xisa [38]

D.the amount of time in sun

8 0

3 years ago

Read 2 more answers

A vector has an x component of 25.0 units and a y

sammy [17]

Here is the answer to your question

8 0

3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

3 years ago